Areas & Volumes of Revolution

∫ f dx for area, π∫ f^2 dx for volume.

The idea

Integration measures more than the area under a curve: the same slicing that built the definite integral also gives the area between two curves and the volume of a solid of revolution.

For $f(x) \ge 0$, the slice at $x$ is a thin rectangle of height $f(x)$, so $\int_{a}^{b} f(x)\,dx$ is the area under the curve.

Between two curves, a vertical slice runs from the lower curve up to the upper one, so its height is the difference of the two function values. Adding the slices gives the area between the curves: $\int_{a}^{b} \big( \text{top} - \text{bottom} \big)\,dx.$

Rotate the region under $y = f(x)$ about the $x$-axis, and the slice at $x$ sweeps out a disc whose radius is the height $f(x)$ of the curve. The disc's face has area $\pi [f(x)]^{2}$, and adding the discs gives the volume of revolution: $V = \pi \int_{a}^{b} [f(x)]^{2}\,dx.$

In every case the method is the same: write down the size of one thin slice, then integrate it across the interval.

Ways to work on it

• Walkthrough. Area under and between curves, and the disk volume formula.
• Practice. Compute the area under a curve as a definite integral.
• Hardest. A volume of revolution as a multiple of π.

Not sure where to start? Take the ten-question placement test.